Question: A particle moves in the $xy$ -plane so that at any time $t\geq 0$ its position vector is $(2t^3+10t,-t^3+2t)$. What is the magnitude of the particle's acceleration vector at $t=1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $5$ (Choice B) B $6\sqrt5$ (Choice C) C $3\sqrt{5}$ (Choice D) D $\sqrt{145}$
Background When solving motion problems, it's important to remember the relationship between the position vector $(x,y)$, the velocity vector $\vec{v}(t)$, and the acceleration vector $\vec{a}(t)$ of the moving particle: $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$ $\vec{a}(t)=\dfrac{d}{dt}\vec{v}(t)=\left(\dfrac{d^2x}{dt^2},\dfrac{d^2y}{dt^2}\right)$ Setting up the math We are given that the particle's position vector is $(2t^3+10t,-t^3+2t)$. We are asked to find the magnitude of the particle's acceleration vector at $t=1$. In other words, we need to find $||\vec{a}(1)||$. Finding $\vec{v}(t)$ $\begin{aligned} \vec{v}(t)&=\left(\dfrac{d}{dt}(2t^3+10t),\dfrac{d}{dt}(-t^3+2t)\right) \\\\ &=(6t^2+10,-3t^2+2) \end{aligned}$ Finding $\vec{a}(t)$ $\begin{aligned} \vec{a}(t)&=\dfrac{d}{dt}\vec{v}(t) \\\\ &=\left(\dfrac{d}{dt}(6t^2+10),\dfrac{d}{dt}(-3t^2+2)\right) \\\\ &=(12t,-6t) \end{aligned}$ Finding $\vec{a}(1)$ $\begin{aligned} \vec{a}({1})&=(12({1}),-6({1})) \\\\ &=(12,-6) \end{aligned}$ Finding $||\vec{a}(1)||$ $\begin{aligned} ||\vec{a}(1)||&=||(C{12},{-6})|| \\\\ &=\sqrt{(C{12})^2+({-6})^2} \\\\ &=\sqrt{180} \\\\ &=6\sqrt5 \end{aligned}$ In conclusion, the magnitude of the particle's acceleration vector at $t=1$ is $6\sqrt5$.